Law of cosines
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==Proof== | ==Proof== | ||
===By the Pythagorean Theorem=== | ===By the Pythagorean Theorem=== | ||
- | An easy way to think of the law of cosines is as an extension of the Pythagorean theorem: | + | An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle: |
::<math>a^{2} + b^{2} = c^{2} </math> | ::<math>a^{2} + b^{2} = c^{2} </math> | ||
- | [[Image: | + | [[Image:Pythagoras_law_of_cosines.jpg]] |
+ | By Pythagorean theorem, we know | ||
+ | ::<math> c^{2} = p^{2} + h^{2}</math> | ||
+ | |||
+ | But <math> p</math> is just some portion of side length <math>a</math> which is <math>r</math> less than the length of <math>a</math>. Substituting the difference gives us, | ||
+ | |||
+ | ::<math> c^{2} = (a-r)^{2} + h^{2}</math> | ||
+ | |||
+ | By Pythagorean theorem, we also know that | ||
+ | |||
+ | ::<math> h^{2} = b^{2} - r^{2}</math> | ||
+ | |||
+ | Substituting the appropriate values gives us, | ||
+ | |||
+ | ::<math> c^{2} = (a-r)^{2} + ( b^{2} - r^{2})</math> | ||
+ | |||
+ | Expanding the squared term gives us | ||
+ | |||
+ | ::<math> c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}</math> | ||
+ | |||
+ | Simplify for | ||
+ | |||
+ | ::<math> c^{2} = a^{2} -2ar + b^{2}</math> | ||
+ | |||
+ | And by the definition of cosine, we know that | ||
+ | |||
+ | ::<math>r=b \cos C</math> | ||
+ | |||
+ | Substituting this value in give us | ||
+ | |||
+ | ::<math> c^{2} = a^{2} -2ab \cos C + b^{2}</math> | ||
+ | |||
+ | or | ||
+ | |||
+ | ::<math> c^{2} = a^{2}+ b^{2} -2ab \cos C </math> | ||
===Using the Distance Formula=== | ===Using the Distance Formula=== | ||
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- | <math> distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}</math> | + | ::<math> \text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}</math> |
<math>c</math> is the distance from <math> A</math> to <math> B</math>. | <math>c</math> is the distance from <math> A</math> to <math> B</math>. | ||
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Substituting the appropriate points into the distance formula gives us | Substituting the appropriate points into the distance formula gives us | ||
- | <math> c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}</math> | + | ::<math> c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}</math> |
Squaring the inner terms, we have | Squaring the inner terms, we have | ||
- | <math> c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}</math> | + | ::<math> c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}</math> |
Since <math> \cos^{2} C + \sin^{2} C = 1</math>, | Since <math> \cos^{2} C + \sin^{2} C = 1</math>, | ||
- | <math> c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}</math> | + | ::<math> c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}</math> |
Square both sides for | Square both sides for | ||
- | <math> c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}</math> | + | ::<math> c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}</math> |
==Example Problem== | ==Example Problem== | ||
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- | <math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math> | + | ::<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math> |
[[Image:SAS triangle.jpg]] | [[Image:SAS triangle.jpg]] | ||
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- | <math> c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ </math> | + | :<math> c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ </math> |
Simplify for | Simplify for | ||
- | <math> c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ </math> | + | ::<math> c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ </math> |
Since <math> \cos 45^\circ = \frac{1}{\sqrt{2}}</math>, substitution gives us | Since <math> \cos 45^\circ = \frac{1}{\sqrt{2}}</math>, substitution gives us | ||
- | <math> c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}}) </math> | + | ::<math> c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}}) </math> |
Simplify for | Simplify for | ||
- | <math> c^{2} =36 + 72 - 72 </math> | + | ::<math> c^{2} =36 + 72 - 72 </math> |
- | <math> c^{2} =36 </math> | + | ::<math> c^{2} =36 </math> |
Taking the square root of both sides gives us | Taking the square root of both sides gives us | ||
- | <math> c =6 </math> | + | ::<math> c =6 </math> |
---- | ---- | ||
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- | <math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math> | + | ::<math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math> |
Substituting in the appropriate side lengths gives us | Substituting in the appropriate side lengths gives us | ||
- | <math> (6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B </math> | + | ::<math> (6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B </math> |
Simplify for | Simplify for | ||
- | <math> 36 (2) = 36 + 36 - 72 \cos B </math> | + | ::<math> 36 (2) = 36 + 36 - 72 \cos B </math> |
- | <math>72 = 72 - 72 \cos B </math> | + | ::<math>72 = 72 - 72 \cos B </math> |
Subtracting <math>72</math> from both sides gives us | Subtracting <math>72</math> from both sides gives us | ||
- | <math>0 = - 72 \cos B </math> | + | ::<math>0 = - 72 \cos B </math> |
Dividing both sides by <math>-72</math> gives us | Dividing both sides by <math>-72</math> gives us | ||
- | <math>0 = \cos B </math> | + | ::<math>0 = \cos B </math> |
Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that | Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that | ||
- | <math>B = 90^\circ </math> | + | ::<math>B = 90^\circ </math> |
And we can find the last angle measure <math>A</math> by subtracting the other two measures from <math> 180^\circ</math> | And we can find the last angle measure <math>A</math> by subtracting the other two measures from <math> 180^\circ</math> | ||
- | <math> 180^\circ - 90^\circ - 45^\circ = 45^\circ</math> | + | ::<math> 180^\circ - 90^\circ - 45^\circ = 45^\circ</math> |
- | <math> A=45^\circ</math> | + | ::<math> A=45^\circ</math> |
[[Image:SAS_solution.jpg]] | [[Image:SAS_solution.jpg]] |
Revision as of 11:25, 3 June 2011
Law of Cosines |
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Law of Cosines
- The law of cosines is a trigonometric extension of the Pythagorean Theorem.
Contents |
Basic Description
The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.When to use it: SAS, SSS.
A More Mathematical Explanation
Proof
By the Pythagorean Theorem
An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle:
By Pythagorean theorem, we know
But is just some portion of side length which is less than the length of . Substituting the difference gives us,
By Pythagorean theorem, we also know that
Substituting the appropriate values gives us,
Expanding the squared term gives us
Simplify for
And by the definition of cosine, we know that
Substituting this value in give us
or
Using the Distance Formula
Let be oriented so that is at the origin, and is at the point.
is the distance from to .
Substituting the appropriate points into the distance formula gives us
Squaring the inner terms, we have
Since ,
Square both sides for
Example Problem
Complete the triangle using the law of cosines.
Solution
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